How to calculate the integral $\int\frac{1}{\sqrt{(x^2+8)^3}}dx$? I need to solve something like this
$$\int\frac{1}{\sqrt{(x^2+8)^3}}dx$$
Wolfram alpha says the solution is $$\frac{x}{8\sqrt{x^2+8}} + c$$
The problem is that the integrand is obtained by the quotient rule:
$$\bigg(\frac{g(x)}{h(x)}\bigg)'=\frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$
$$\bigg(\frac{x}{8\sqrt{x^2+8}}\bigg)'=\frac{1}{8}\cdot\frac{\sqrt{x^2+8}-\frac{x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{\frac{x^2+8-x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{8}{\sqrt{(x^2+8)^3}}=\frac{1}{\sqrt{(x^2+8)^3}}$$
It there a way to extract the solution from these types of integrals which argument is born from the easy quotient rule?
 A: substitute $$x=2\sqrt{2}\tan(u)$$ then we get
$$dx=2\sqrt{2}\sec^2(u)du$$
A: 
It there a way to extract the solution from these types of integrals which argument is born from the easy quotient rule?

None that I know of. You just see it or maybe make a lucky substitution. 
Alternatively, you can take a look at Euler substitutions, which will work in this case and can help for quite a lot of hand-made integrals for homeworks.
A: I don't know of any general formula for the quotient rule.
But for an integral of the form
$$\int \frac{dx}{\left(ax^n+b\right)^{k}}$$
You can try the following substitutions:


*

*If $n=1$, put $u=ax+b$. ($n=0$ is trivial.)

*If $n=2$, put $x=\left(\frac{b}{a}\right)^\frac{1}{n} tan u$.

*If $n\ge 3$, put $x=\left(\frac{b}{a}\right)^\frac{1}{n} u$ and then factorise the polynomial in the denominator into linear and quadratic terms and then, after using the method of partial fractions to separate the terms, apply 1 and 2 as above.


This is just a general procedure, and may not be applicable always. Some problems will possess solutions obtained only by special methods.
Hope this helps.
A: As mentioned in another answer, the solution to the given integral can be found by making a substitution with a trigonometric function, then integrating, calculating, substituting back and simplifying.
For a general integral of the form
$I =\displaystyle \int \frac{1}{\left(a x^n+b\right)^p} \, dx$ ,
the solution can be obtained with the help of Mathematica in terms of a special function:
$\displaystyle I = x \left(a x^n+b\right)^{-p} \left(\frac{a x^n}{b}+1\right)^p \, _2F_1\left(\frac{1}{n},p;1+\frac{1}{n};-\frac{a x^n}{b}\right)+C$,
where $\, _2F_1(a,b;c;z)$ is the Gaussian hypergeometric function.
The integral in this question is the special case where
$a =1, b = 8, n = 2,p =3/2$
A function can be defined with Mathematica:
solh[a_, b_, n_, 
   p_] = (x*(1 + (a*x^n)/b)^p*
     Hypergeometric2F1[1/n, p, 1 + 1/n, -((a*x^n)/b)])/
       (b + a*x^n)^p; 

Then taking the special values for the integral in this question:
FullSimplify[solh[1, 8, 2, 3/2]]

It can be verified that the solution is
$\displaystyle \frac{x}{8\sqrt{x^2+8}} + constant$
